Are there continuous functions for which the epsilon-delta property doesn't hold? The standard definition of continuity is as follows:
A function is continuous if $$\forall \varepsilon > 0\  \exists \delta > 0\ \text{s.t. } 0 < |x - x_0| < \delta \implies |f(x) - f(x_0)| < \varepsilon $$
This may sound silly, but is the converse true? In other words, is there an example of a function which is continuous, but for which this property doesn't hold? I thought I remembered reading something about the exponential function being an example of a continuous function for which the epsilon-delta implication is false, since it gets arbitrarily steep as x approaches infinity, but I could be wrong. Thanks!
 A: The definition of a continuous function is that it satisfies the $\epsilon-\delta$ property. Definitions, by convention, work both ways. Hence, we would say that any function satisfying the $\epsilon-\delta$ property is continuous by definition.
I should add that the $\epsilon-\delta$ property, is something that would hold only in metric spaces i.e. where you have a distance function between points. In a general topological space (which need not have a distance function on it!), the definition of continuity of a function $f$ between topological spaces $X$ and $Y$ , is that the the pre-image of an open set in $Y$ must be open in $X$.
As it turns out, if $X$ and $Y$ are metric spaces, then this definition is equivalent to the $\epsilon-\delta$ property, which means that on metric spaces you can work with either definition and your job will be done.
Hence, continuity is equivalent to the $\epsilon-\delta$ property wherever this property is applicable.
A: The statement you made is the definition of continuity (in a metric space, where the notion of distance or $|x-x_0|$ makes sense).  So it works both ways:  If the function is continuous, it meets the definition and if it meets the definition then it is continuous.
You are encountering the idea of a function being "uniformly continuous."  In the definition you wrote, there is nothing said about whether then same $\delta(\epsilon)$ can work irrespective of the point $x_0$.  Your exponential example is indeed continuous, but it is not uniformly continuous, because $\delta$ has to depend both on $\epsilon$ and $x_0$.  
In contrast, a function like $\frac{x^3}{1+x^2}$ is uniformly continuous, because for any given $\epsilon > 0$ you can find a $\delta$ such that that defining condition holds for every real $x_0$.
A: This article from wikipedia may also look interesting. Especially this part:

In mathematics, a definition is used to give a precise meaning to a new term, instead of describing a pre-existing term. Definitions and axioms are the basis on which all of mathematics is constructed.

And:

An Intensional definition, also called a connotative definition, specifies the necessary and sufficient conditions for a thing being a member of a specific set. Any definition that attempts to set out the essence of something, such as that by genus and differentia, is an intensional definition.

A: Since it's a definition, your first "if" should really be read as an "if and only if". So the converse you ask about always holds.
